An object speeding up, an object slowing down, and an object changing direction are all examples of acceleration.

Physics

1 answer:

emmainna [20.7K]2 years ago

6 0

<h2>Answer: True</h2>

When an object is speeding up, the acceleration is in the same direction as the velocity. Thus, this object has a positive acceleration. In Example B, the object is moving in the negative direction (i.e., has a negative velocity) and is slowing down.

You might be interested in

In a particular beam of radiation, which is traveling in a vacuum, the amounts of energy per second at an ultraviolet wavelength

Darina [25.2K]

Answer: d)

Explanation: In order to justify the answer we have to consider that the energy of photons directely depent on the frequency so the energy is inverselly dependent of the wavelegth.

If both beams have the same power, this means Energy/time so the number of photons per second must be different. As consequence a) is wrong as b) since it is not posible since UV photon have more energy that IR photons. c) It is no necessary know the frequency since the wavelength is related in the form:

c=λν c is the speed of light, λ the wavelegth and ν the frequency.

d) Certainly will be more more IR photons than UV photons to get the same beam power.

8 0

3 years ago

Which group of elements are shiny, opaque, and have a high melting point?

vlabodo [156]

The answer is A. Metals

3 0

3 years ago

A 30 g bullet moving a horizontal velocity of 500 m/s comes to a stop 12 cm within a solid wall. (a) what is the change in the b

Sidana [21]

M = 30 g = 0.03 kg, the mass of the bullet
v = 500 m/s, the velocity of the bullet

By definition, the KE (kinetic energy) of the bullet is
KE = (1/2)*m*v²
= 0.5*(0.03 kg)*(500 m/s)² = 3750 J
Because the bullet comes to rest, the change in mechanical energy is 3750 J.

The work done by the wall to stop the bullet in 12 cm is
W = (1/2)*(F N)*(0.12 m) = 0.06F J

If energy losses in the form of heat or sound waves are ignored, then
W = KE.
That is,
0.06F = 3750
F = 62500 N = 62.5 kN

Answer:
(a) 3750 J
(b) 62.5 kN

7 0

3 years ago

A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the

Mamont248 [21]

Answer:

The minimum coefficient of friction required is 0.35.

Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

$-F_{f} + F = 0$